Do polynomials with complex coefficients need to have complex conjugate roots? From the Complex conjugate root theorem we get that if a polynomial in one varaible with real coefficients has as solution  $a + bi$ , than $a-bi$ must also be a solution...however, what happens if the coefficients are not real? How do we solve then? 
 A: If a polynomial has complex coefficients, then its roots could be any complex numbers, and they are not guaranteed to occur in conjugate pairs. For example, we can obtain a polynomial with roots at $i$ and $1+i$, simply by writing 
$p(x) = (x-i)(x-(1+i)) = x^2 -(1+2i)x + (-1+i)$.
This can be solved with the quadratic formula, just like a polynomial with real coefficients.
As the degree of the polynomial increases, solving it gets harder. However, you can always reduce the problem to solving two real polynomials in two variables, one for the real part and one for the imaginary part. For example, if we want to solve:
$x^3 - ix + 2i = 0$,
we can start by writing $x=a+bi$. Substituting this into our equation, we get:
$
(a+bi)^3 - i(a+bi) + 2i = 0 \\
a^3 + a^2bi - ab^2 - b^3i -ai + b + 2i =0
$
Now you want to separate real and imaginary parts:
$(a^3-ab^2+b) + (a^2b-b^3-a+2)i = 0$
Since the real and imaginary parts of $0$ are both $0$, this gives us two equations in the two variables $a$ and $b$:
$a^3-ab^2+b=0\\
a^2b-b^3-a+2=0$
You could use the quadratic formula now to solve for $b$ in the first equation, substitute that value into the second equation, and have a single equation in $a$. Since $a$ is a real variable, you can use all the usual techniques to solve for it, and then back-substitute to get $b$.
If the degrees are higher still, then your system of two polynomials in two variables could be kind of hard. There are calculations with "Groebner bases" that help with elimination in these cases, but eventually, you have to solve a high-degree polynomial for one real variable.
